Let $g(x)=xe^{3x}$. What is the absolute minimum value of $g$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac{1}{3e}$ (Choice B) B $\dfrac{3}{e^3}$ (Choice C) C $-\dfrac{1}{e^3}$ (Choice D) D $g$ has no minimum value
Answer: Let's first find the relative extremum points of $g$, and then consider them along with the function's end behavior in both directions. We start with finding the critical points of $g$. The derivative of $g$ is $g'(x)=e^{3x}(1+3x)$. $g'(x)=0$ for $x=-\dfrac{1}{3}$. $g'$ is defined for all real numbers. Therefore, our only critical point is $x=-\dfrac{1}{3}$. Our critical point divides the function's domain (which is all real numbers) into two intervals: $\llap{-}6$ $\llap{-}5$ $\llap{-}4$ $\llap{-}3$ $\llap{-}2$ $\llap{-}1$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $(-\infty, -\frac{1}{3})$ $(-\frac{1}{3}, \infty)$ $-\frac{1}{3}$ Let's evaluate $g'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $g'(x)$ Verdict $(-\infty,-\dfrac13)$ $x=-1$ $g'(-1)=-\dfrac{2}{e^3}<0$ $g$ is decreasing $\searrow$ $(-\dfrac{1}{3},\infty)$ $x=0$ $g'(0)=1>0$ $g$ is increasing $\nearrow$ Let's imagine ourselves walking on the graph of $g$, starting all the way to the left (from $-\infty$ ) and going all the way to the right (until $+\infty$ ). According to the table, we will start by going down and down until we reach $x=-\dfrac{1}{3}$. Then, we will be forever going up. Therefore, $g$ must obtain its absolute minimum value at $x=-\dfrac{1}{3}$. We are asked to find that minimum value, which is $g\left(-\dfrac{1}{3}\right)=-\dfrac{1}{3e}$. In conclusion, the absolute minimum value of $g$ is $-\dfrac{1}{3e}$.